Electric field calculation

Quote:

Originally Posted by free_to_fly

I'm stuck on this question:
A sphere of radius R is charged to a potential V far from any other bodies and is inserted inside the inner conductor of the cylindrical capacitor without touching it.

First consider the sphere with charge Q which starts out physically remote from anything else. We have by Gauss:

$Q=\oint_S\epsilon \underline E\cdot \underline{dS}$

Where S is any closed surface enclosing the sphere. The problem has spherical symmetry so if we choose a sphere of radius r for the surface S then the E field is normal to the surface and the same magnitude everywhere. This makes the integral easy to solve:

$Q=\oint_S\epsilon \underline E\cdot \underline{dS}= \epsilon E*(4 \pi r^2)$ or $E=\frac Q{4 \epsilon \pi r^2}$

Now the voltage at the point of infinity is zero so we can find the voltage on the surface of the sphere by integrating E from infinity to the surface of the sphere:

$V=-\int_\infty ^R E dr = -\int_\infty ^R \frac Q{4 \epsilon \pi r^2}=-\frac Q{4 \epsilon \pi} \int_\infty ^R \frac 1 {r^2}=\frac Q{4 \epsilon \pi R}$

or $Q = 4 \epsilon \pi R V$

Now once the sphere is placed inside your capacitor the plates of the capacitor force the E field to be uniform between the concentric plates. Therefore, in this area the field will have cylindrical symmetry. We use Gauss again only this time we choose a cylinder radius r, length l for our integrating surface:

$Q=\oint_{S'}\epsilon \underline E\cdot \underline{dS}=2 \pi rl \epsilon E$ or $E=\frac Q {2 \pi \epsilon rl}$

Substituting for Q gives:

$E=\frac {4 \epsilon \pi R V} {2 \pi \epsilon rl} =\frac{2RV}{rl}$